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gayaneshka [121]
3 years ago
13

Analyze the following chemical combinationsa) MgOb)CaCl2c) Na2O​

Chemistry
1 answer:
Rashid [163]3 years ago
8 0
The answer is B for sure
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How many grams of phosphorus are in a sample of Ca3(PO4)2 that contains 50.0 g of calcium? Answer: ________ grams of phosphorus.
wel

<u>Answer:</u> The mass of phosphorus that is present for given amount of calcium is 28.53 g.

<u>Explanation:</u>

We are given:

Mass of calcium = 50 grams

The chemical formula of calcium phosphate is Ca_3(PO_4)_2

Molar mass of calcium = 40 g/mol

Molar mass of phosphorus = 31 g/mol

In 1 mole of calcium phosphate, 120 grams of calcium is combining with 62 grams of phosphorus.

So, 50 grams of calcium will combine with = \frac{62}{120}\times 50=28.53g of phosphorus.

Hence, the mass of phosphorus that is present for given amount of calcium is 28.53 g.

6 0
3 years ago
A(n) ____________________ is a group of two or more atoms that are held together by chemical bonds. *
Kryger [21]

Answer:

A molecule is a group of two or more atoms that are held together by chemical bonds.

Explanation:

6 0
3 years ago
How do you think the crocodile might affected if it wandered near a strong magnet?
vodomira [7]
According to florida wildlife group who experimentally tape magnets to crocodile heads to disrupt their homing ability so they don't wander into residential areas
5 0
3 years ago
Consider the following balanced equation:
algol [13]

Moles of PF₃ : 4

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

\tt P_4(s)+6F_2(g)\rightarrow 4PF_3(g)

1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)

Limiting reactant : the smallest ratio (mol divide by coefficient)

P₄ : F₂ =

\tt \dfrac{1.25}{1}\div \dfrac{6}{6}=1.25\div 1\rightarrow F_2~limiting~reactant(smallest~ratio)

mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :

\tt \dfrac{4}{6}\times 6~moles=4~moles

8 0
3 years ago
If the concentration of the HCl used in your titration was 1.00 M and you used 23.68 mL to reach the endpoint, calculate the con
drek231 [11]

Answer: The concentration of the OH-, CB = 0.473 M.

Explanation:

The balanced equation of reaction is:

2HCl + Ca(OH)2 ===> CaCl2 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = 1M

CB is the concentration of base = to be calculated

VA is the volume of acid = 23.65 ml

VB is the volume of base = 25mL

Substituting

1×23.65/CB×25 = 2/1

Therefore CB =1×23.65×1/25×2

CB = 0.473 M.

6 0
3 years ago
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